Bouncing balls in a heptagon · deepseek-v4-flash-0731

Let me build a 2D physics simulation of numbered balls inside a rotating heptagon. Let me plan the implementation carefully.

Key requirements:
1. 20 balls numbered 1-20 in compact non-overlapping arrangement near center, fall under gravity
2. Rotating heptagon (7 sides), one revolution every 5 seconds, always fully visible
3. Ball-ball and ball-wall collision detection, moving-wall velocity in bounce, damping/friction
4. Angular motion, numbers rotate with balls
5. Reasonable sizes/constants, stable
6. Pause/Resume, Reset

Let me think about the physics.

For a rotating heptagon, the walls move. When a ball collides with a wall, we need to account for the wall's velocity.

The heptagon is a regular polygon rotated around its center. Let me define the world coordinates. The heptagon center is at canvas center. The rotation angle θ(t) = ω*t, where ω = 2π/5 rad/s.

For each side of the heptagon, I need to compute collision. A regular heptagon with circumradius R has vertices at angles. Each edge is between two consecutive vertices.

Hmm, this is tricky. Collision with an arbitrary convex polygon that's rotating. A robust approach: for each edge, compute the closest point on the edge to the ball center, and check if distance < ball radius.

But dealing with moving walls: the wall velocity at the contact point. For rotation, velocity of a point on the rigid body is v = ω × r where r is the vector from center of rotation to the point. For 2D, v = ω_perp applied: v = (-ω*y, ω*x) where (x,y) is relative to rotation center.

Let me set up coordinates with center of rotation at the heptagon center C.

Actually, let me make the enclosure fixed in world space rotating around world center O. Balls live in world space. The heptagon geometry is the polygon, rotating. 

For collision detection between a ball and the rotating heptagon:

Let me represent each edge. The polygon vertices in world coords:
vertex[i] = O + R * (cos(θ0 + θ + i*2π/7), sin(θ0 + θ + i*2π/7))

For each edge (v_i, v_{i+1}), compute closest point from ball center to segment. If distance < ball radius, we have a collision.

The wall velocity at contact point: v_wall = ω_perp × (contact - O). In 2D with rotation about O, v_wall = (-θdot*(contact.Y - O.Y), θdot*(contact.X - O.X)).

Wait, for rotation about origin, angular velocity vector points out of screen (positive z). Velocity = ω × r where in 2D cross product gives v = (-ω*ry, ω*rx). Where r = contact - O.

So v_wall_x = -ω * (contact.Y - O.Y)
v_wall_y =  ω * (contact.X - O.X)

Relative velocity = ball_v - wall_v at contact.

Collision resolution for ball-wall:
- Compute normal n (unit normal pointing from wall into interior where ball should be, i.e., from contact outward to ball center, pointing away from wall toward ball).
- Project relative velocity onto normal: vn = dot(rel_v, n)
- If vn < 0 (approaching), apply impulse: separate ball.

The normal for restitution: rel_v_n_after = -e * rel_v_n_before. So change in normal velocity = -(1+e)*vn.

Ball mass m, so impulse magnitude j = -(1+e)*vn (for unit... well impulse on ball). New ball velocity = ball_v - n*( (1+e)*vn )? Let me be careful.

Actually the standard: j = -(1+e) * (vn) where vn = dot(v_rel, n) computed with v_rel = velocity of ball minus velocity of wall. Then new v_ball = v_ball - (j / m) * n? Let's define n pointing from wall toward ball (outward normal). Approach when rel velocity has negative component along n.

v_new_ball = v_ball + (1+e)*(-vn)/1 * n = v_ball - (1+e)*vn * n. Since vn negative, that adds velocity away from wall. Good.

Wait let me define j_n = -(1+e)*vn (positive when vn negative). Then Δv = (j_n/m)*n.

Moving the ball out of penetration: set ball center to contact point + n*(radius + small epsilon). Actually for wall collision, ball is in interior but penetrated; we push it back along n to be just touching: center = contact + n*(radius + epsilon). contact is the closest point on the segment inside the wall.

Friction: tangential relative velocity vt = rel_v - n*vn. Apply friction reducing tangential relative velocity by friction coefficient * |normal impulse|. Set v_new tangential = rel after... This is complex. Let me do: after handling normal (position + normal velocity), apply tangential friction to change ball's velocity such that relative tangential velocity is reduced.

Let me implement step by step per collision.

For angular motion with friction: friction force at contact produces torque on ball, causing spin. The number rotates with ball. Let me track ball.angle and ball.angularVel.

Simplification: when there's a collision (wall or ball), also impart some angular velocity based on tangential relative velocity, and the angular velocity decays via friction damping.

I think a reasonable approach:

Ball-ball collision: elastic-like with some restitution. Handle momentum exchange. Also impart spin based on relative tangential velocity.

Let me keep it reasonably simple but functional.

Let me structure the physics update:

For each frame dt:
1. Apply gravity: v += g*dt; angle += angVel*dt (or update based).
2. Integrate position: pos += v*dt.
3. Collision detection/response for walls, then balls. Possibly iterate a few times for stability at small dt.

To keep stable, use fixed timestep and spatial hash or simple O(n^2) for 20 balls — trivial.

Let me pick values in a coordinate system where canvas is maybe 700x700 or something. Let me use logical units and scale.

Canvas size: let me make it square, e.g., 900x900 or adaptive. Let me use a fixed internal resolution and scale via CSS. Actually let me just set canvas width/height to something and center heptagon.

Let me use canvas of 800x800. Heptagon circumradius R = 300, center at (400, 400).

Ball radius: with 20 balls in a compact group... Let's say ball radius ~28 px but that might be big relative to 300 R heptagon. Inradius of heptagon = R*cos(π/7) ≈ 300*0.9009 ≈ 270. So interior radius ~270. If balls radius 26, they fit.

Actually for a compact arrangement of 20 balls near center that then fall, we arrange them in a hexagonal/triangular lattice packing. Could arrange in rows like a triangle: 1+2+...+n = 20 → n=5 gives 15, n=6 gives 21. So near-natural triangle of 21 minus one. Let me just hand-place a compact cluster.

Let me arrange: rows of increasing balls but offset. I'll just generate a compact packing.

Compact packing for equal circles diameter D = 2r. Place in hex lattice with spacing exactly 2r (touching). Rows alternate offset by r. Let me do a triangular arrangement:

Row 0: 1 ball
Row 1: 2 balls
Row 2: 3 balls
Row 3: 4 balls
Row 4: 5 balls
Row 5: 5 balls
Total = 20. Good, that's 20 balls in a roughly triangular/hex cluster, compact and non-overlapping if spacing = 2r.

Position: center the cluster at heptagon center. For each row y, x centered.

Let ball numbering: ball number = index+1.

Placement: 
Row i (0-indexed) has count c. Each row's balls at y = cy + i*rowH, where rowH = spacing * sqrt(3)/2 (vertical distance between rows in hex lattice). x centered: x = cx - (c-1)*spacing/2 + j*spacing, offset alternate rows by spacing/2.

With spacing = 2r + tiny gap. Let r=27, spacing=55. Row height = 55*0.866 = 47.6.

Rows heights: total height 6 rows * 47.6 ≈ 286. Cluster width max 5*55=275. Fits within inradius 270 roughly, and near center. Good.

Ball radius 27 might make cluster near touching the walls when falling. Interior inradius 270, cluster radius ~143. Fine.

Let me set constants:
- Gravity g = 900 px/s² (in canvas units)
- dt = 1/120 for physics substeps
- restitution e = 0.6
- friction coefficient ~0.3
- angular damping

Rotation: ω = 2π/5 rad/s, θ(t) = ω*t.

Let me also handle collision detection carefully with moving walls.

Physics loop with fixed substeps: use accumulator. For each animation frame (requestAnimationFrame), compute real dt (clamped), accumulate, run fixed steps at dt=1/240 maybe for stability. 20 balls, O(n²)=400 pair checks, ~few iterations, trivial performance. Use dt=1/240 might be overkill; 1/120 fine.

Actually for stable with walls and ball bouncing, substep dt of 1/120 with several constraint iterations is fine.

Let me implement collision response:

Wall nearest point function: For segment P0P1 and point P, find closest point. Standard.

For each ball, for each edge:
- compute closest point on segment to ball center.
- dist² < (radius)² → collide.
- normal n = (ball.center - closest)/|...| (pointing from wall to ball). If closest==center (degenerate), skip.
- penetration = radius - dist.
- Push ball out: center = closest + n*(radius + 0.5*penetration)? Actually to fully resolve: move center along n so distance becomes radius: center = closest + n*radius.
- Wall velocity at closest point: vWall = ω_vec cross (closest - O) = (-ωy, ωx).
- rel_v = v_ball - vWall.
- vn = dot(rel_v, n).
- if vn < 0: apply restitution impulse: v_ball += n * (-(1+e)*vn).
- Friction: after normal, compute tangential. vt = rel_v - n*vn (this is current relative tangent). Actually we want to reduce tangential relative motion. 
  Friction reduces relative tangential velocity. Impulse-based friction: target reduction factor. Let me compute: after normal response, recompute rel_v2 = v_ball_new - vWall. tangential component vt2 = rel_v2 - n*(dot(rel_v2,n)). Apply friction: reduce vt2 by factor fc. 
  
  To also convert some tangential linear motion into angular spin: the tangential relative velocity (ball surface slipping vs wall) causes ball to spin. For ball spinning on wall, we can compute angular acceleration from friction.

Let me keep friction application: new tangential relative velocity vt2' = vt2 * (1 - µ). And add to ball angular velocity based on change in tangential velocity: Δω = Δv_tangential / radius. Hmm direction: Δv_tangential (change in ball's linear tangential velocity) relates to angular.

Actually the ball's tangential contact point velocity = v_ball + ω_cross... For a ball, surface velocity at contact = v_ball_t - ω*radius (for rolling). Friction accelerates rotation to reduce slip.

Let me implement: 
- Extract tangential unit t = (-n.y, n.x) or choose based on relative motion direction.
- tangential relative velocity at contact:
  v_slip = v_ball_t - ω × ... In 2D, contact point of ball at bottom (in direction -n from center): point P_contact = center - n*radius. Velocity of that material point = v_ball + ω_ball_z × (P_contact - center). (P_contact - center) = -n*radius. Cross in 2D: ω × r = (-ω*r_y, ω*r_x). r = -n*radius. So surface velocity = v_ball + (-ω*(-n_y*radius), ω*(-n_x*radius)) = v_ball + (ω*n_y*radius, -ω*n_x*radius).
  
  Hmm getting complicated with signs. Let me just use a practical model:

Ball angular velocity updates from friction torque: 
τ from friction force F_friction applied at contact (radius from center = r·n^contact). Torque scalar = cross(r_c, F) where r_c = P_contact - center = n*radius (pointing from center to contact, which is -n outward... wait contact is at the point of wall contact, which is toward the wall, i.e., in the -n direction from center since n points away from wall). So r_c = -n*radius.

This is getting detailed. Let me simplify with a reasonable approximate model that produces visually spinning balls:

When a collision occurs, compute relative tangential velocity at the contact. Then:
- Apply friction to linear tangential velocity (reduce relative slip).
- Set/increment the ball's angular velocity to a value consistent with the change.

Simplest robust: on each collision, after computing tangential relative velocity component v_t (some scalar along tangent), set ball angular velocity toward a spin that matches friction. 

Let me do this concrete approach:
1. Compute normal response (position + normal velocity restitution).
2. Compute tangential unit vector t, and tangential relative slip velocity v_s = dot(rel_v, t) (after normal update).
3. The desired friction causes the contact surface of the ball to not slip: ball surface velocity at contact in tangential direction should become 0. Ball surface velocity_t at contact (relative) includes linear term and spin term: v_surface_ball = v_s_lin_ball + ω*radius (sign). Hmm.

Let me just directly model: friction force f = µ * |normal impulse| (Coulomb). Direction opposes slip. Normal impulse magnitude = (1+e)*|vn| (impulse per unit mass momentum change). Change in ball linear tangential velocity magnitude = f (impulse per mass, since j/m, m normalized to 1). Change in angular velocity = (f * radius) / I_per_mass where I = (2/5) m r² for solid sphere → I/m = (2/5)r². For a disk, (1/2) r². Use I/m = 0.4*r² (sphere) or 0.5*r² (disk). Use disk ~ r²/2.

Δω = (f_impulse * radius) / (0.5 r²) = 2*f_impulse/r.

And direction: friction torque acts to reduce slip. Sign of Δω relative to slip direction.

Given complexity, let me implement a robust standard approach for ball on a surface (wall):

After resolving normal collision:
- rel_v computed BEFORE normal (relative velocity), vn = dot(rel_v,n), rel_v_t = rel_v - n*vn.
- tangent direction t = rel_v_t normalized if nonzero.
- slip speed s = |rel_v_t|.
- normal impulse magnitude Jn = (1+e)*vn_neg where... let me define vn (approaching, negative). Normal impulse per mass j_n = (1+e)*(-vn).
- friction: static-ish: reduce slip. Maximum friction impulse j_f = µ * j_n. Apply reducing ball tangential: Δv_t = j_f (in -t direction = oppose relative motion of ball). But careful: rel_v = v_ball - v_wall. To reduce relative slip, we change v_ball tangent by -j_f*t (opposing ball's relative motion). 
- So new v_ball_t = v_ball_t_old - j_f*t? Actually t is direction of rel_v_t (ball moving relative to wall along t). To reduce slip, decrease ball's velocity along t by j_f: v_ball_t_new = v_ball_t - j_f*t.

Wait but friction impulse on ball should oppose slip, which is direction of rel_v (ball relative to wall). So yes Δv_ball = -j_f * t_hat.

- Then angular: ball spins due to friction. The torque from friction (applied at contact, force pointing -t_hat, at radius from center) causes rotation. Δω such that... The friction force -j_f*t acting at contact point at position (center - n*r) relative... The resulting change in spin: Δω = -(something).

I'll set angular based on coupling: the surface's tangential velocity change due to spin = +ω*r (for bottom, with proper sign). Friction imparts spin. Let me use: Δω = (-j_f/r) * sign... 

You know, let me define sign conventions concretely and just make it visually work:

Let me define for a wall collision, normal n points from wall outward (into the enclosure interior), toward the ball. 

Tangent t = (n.y, -n.x). The relative velocity rel_v = v_ball - v_wall. The tangential part rel_v_t = rel_v - dot(rel_v,n)*n. Let me write rel_v_t as a scalar along t: since rel_v_t is perpendicular to n, and t is the perpendicular direction, rel_v_t = (dot(rel_v_t, t)) * t. Let s = dot(rel_v, t) (scalar relative tangential speed along t).

Ball surface velocity at the contact point relative to its center: the contact point on ball is at center - n*radius (toward wall). For 2D rotation positive counterclockwise (z out of screen), the velocity of a point at offset r_vec from center due to spin ω (CCW positive) is v = ω × r_vec = ω * (-r_y, r_x) [cross in 2D with ω as scalar z].

r_vec = -n*radius. So spin surface velocity = ω * (-(-n_y*radius), (-n_x*radius)) = ω*radius*(n_y, -n_x) = ω*radius*t.

So spinning the ball CCW (ω>0) adds surface velocity +ω*radius in the +t direction at the contact.

Total relative slip at contact in t direction: s_contact = dot(v_ball, t) + ω*radius - dot(v_wall, t). (ball surface velocity along t minus wall surface velocity along t.)

For pure rolling (no slip): s_contact = 0 → dot(v_ball,t) + ω*radius = dot(v_wall,t).

The friction tends to drive s_contact → 0. 

Impulse approach for the tangential direction including spin (rolling without slipping constraint): 
To remove slip entirely in one step (like a kinematic coupling), we could solve: after impulse, s_contact' = 0. Let v_ball change by Δv along t, and ω change by Δω.
s_contact' = (dot(v_ball,t)+Δv) + (ω+Δω)*radius - dot(v_wall,t) = 0.
Friction impulse relation: Δv = -j_f (per mass), and Δω = (torque)/(I/m) = (j_f*radius)/(I/m) but sign. Torque from friction force at contact: friction force along -t? 

Friction force opposes slip. If s_contact > 0 (surface slipping in +t direction), friction acts -t direction on the ball at contact. Force F = -j_f*t. Torque = r_c × F where r_c = contact - center = -n*radius. r_c × F (z-component) = r_c.x*F.y - r_c.y*F.x. With F = -j_f*t = -j_f*(n.y,-n.x), and r_c=-n*radius. This cross product... let me just compute:

n=(nx,ny). t=(ny,-nx). r_c = -radius*(nx,ny). F = -j_f*(ny,-nx) = (-j_f*ny, j_f*nx).
Torque_z = r_c.x*F.y - r_c.y*F.x = (-radius*nx)*(j_f*nx) - (-radius*ny)*(-j_f*ny) = -radius*nx²*j_f - radius*ny²*j_f = -radius*j_f*(nx²+ny²) = -radius*j_f.
So torque = -radius*j_f. Δω = torque/(I/m) = -(radius*j_f)/(0.5 radius²) = -2*j_f/radius.

So Δω = -2*j_f/radius. Given slip s_contact>0 friction -t, j_f>0, Δω<0 decreases ω. Reasonable.

Now solve for j_f to zero slip:
Δv = -j_f. Δω = -2 j_f/r.
0 = s_contact_old + Δv + Δω*r = s_contact_old - j_f - 2*j_f = s_contact_old - 3 j_f.
j_f = s_contact_old/3.

But also limit by Coulomb: j_f_max = µ * j_n. So j_f = min(s_contact_old/3, µ*j_n). And s_contact must have opposite? Only apply if s_contact_old has sign such that friction reduces magnitude. Apply reduction of |s_contact|. Actually j_f from formula reduces slip toward zero by amount 3*j_f. If s_contact_old/3 is large but limited by Coulomb, fine.

This is a clean model. Let me implement for wall collision. For ball-ball collision, do similar (set I for equal balls).

This gives spinning due to friction. Plus I'll add general angular damping so they slow when not colliding.

Actually, to make spin visually clear, also ensure numbers rotate. Track ball.angle += ball.angVel*dt always, and angVel decays by damping factor. During collisions angVel changes per friction model.

I think this is good. Let me also consider that when a ball rests on the rotating wall, friction makes it roll (pick up spin) which naturally demonstrates. Good.

Now, the wall collision detection for a point inside a convex rotating polygon. Edge normal n should point inward (into enclosure) so we push ball inward. My formula uses n = (ball.center - closest)/|...|. Since closest is on wall and ball is interior, (center-closest) points into interior (toward center). Good, that's outward normal (away from wall into ball region = interior). Wait n points from wall toward ball = interior direction. Yes that's the outward normal of the wall from the solid's perspective... The enclosure is like a container; n points from wall into the container interior. Good, that's the direction we push ball. Correct.

But when ball is exactly at vertex (closest = vertex), normal may point along the bisector, fine.

Potential issue: a ball could be pushed outside if penetration large. Use substeps to keep velocities bounded and it's fine.

Edge case: ball center near a vertex—closest point on both adjacent edges is the vertex; collision checks for both edges will trigger with n from vertex; apply twice—may double-correct but acceptable.

Let me also ensure the heptagon is fully visible. Canvas 800x800, center (400,400), R=300 → max distance from center = 300 < 400. Good. Add margin.

Let me also draw the heptagon walls (outline) and maybe slight fill. Number the balls (draw number rotating with ball). Draw a direction marker on wall maybe.

UI: buttons Pause/Resume and Reset. Also maybe show elapsed sim time and rotation angle.

Let me write the code.

Let me define coordinate: world where heptagon rotates about center O = (cw/2, ch/2). θ(t) starts at some θ0. On reset, restore θ0.

Wait "Reset restores the initial ball arrangement and wall orientation." So initial arrangement positions balls and wall orientation θ0.

Let me compute vertex angles: vertex i angle = θ + i*(2π/7). Let me choose θ0 = 0 initially, but to look nice maybe rotate a bit. π/7? Let me use θ0 = -90° offset so a vertex at top? Doesn't matter. I'll choose vertices such that the polygon is symmetric with a flat-ish top or vertex top. Regular heptagon has no opposite parallel. Let me just use θ(t). 

Let me code.

Physics step pseudocode:

let G = 980; // px/s^2
let dt = 1/120;

function stepWorld(dt, simTime){
  // update rotation
  // For each ball: apply gravity, integrate, update angle
  for ball: 
     ball.v.y += G*dt;
     ball.pos += ball.v*dt;
     ball.angle += ball.angVel*dt;
     ball.angVel *= (1 - angDamping*dt) ; // decay
     // clamp speeds
  // wall collision
  for each ball, for each edge:
     detect & resolve (single pass)
  // ball-ball
  several iterations of pair resolution
}

Actually need multiple passes for ball-wall and ball-ball to avoid overlap and settle stacks. Let me do a few iterations (e.g., 2-3) combining detection each iter.

For stability with stacked balls and moving walls, small dt (1/120) and 2 iterations each frame is usually fine.

Let me also add clamping of velocity to prevent blow-up, e.g., if speed > maxSpeed, scale down. Shouldn't happen with dt small.

Let me code the wall edge geometry each physics step (recompute vertex positions based on current θ). 

Wall collision resolve function:

function wallCollide(ball, p0, p1, O, dt){
  const closest = closestOnSegment(ball.pos, p0, p1);
  const dx = ball.pos.x - closest.x, dy = ball.pos.y - closest.y;
  const d2 = dx*dx+dy*dy;
  if(d2 >= R*R) return; // R ball.radius
  if(d2 === 0) return; // center on segment
  const d = Math.sqrt(d2);
  let nx = dx/d, ny = dy/d;
  // push out
  ball.pos.x = closest.x + nx*(R+0.01);
  ball.pos.y = closest.y + ny*(R+0.01);
  // wall velocity at closest
  const rx = closest.x - O.x, ry = closest.y - O.y;
  const vwx = -omega*ry, vwy = omega*rx; // omega radi per sec
  const relx = ball.v.x - vwx, rely = ball.v.y - vwy;
  const vn = relx*nx + rely*ny;
  if(vn < 0){
    const jn = (1+e)*(-vn);
    ball.v.x += jn*nx;
    ball.v.y += jn*ny;
    // friction with spin
    // t = (ny,-nx), slip scalar
    const tx = ny, ty = -nx;
    const srel = relx*tx + rely*ty; // relative surface linear along t
    // contact slip includes spin
    const s_contact = srel + ball.angVel*R;
    // friction: drive to zero, limit coulomb
    const jf = Math.min(Math.abs(s_contact)/3, mu*jn, MAXFRIC) * sign(s_contact);
    // Apply friction to ball velocity along -t? 
    // Formula: Δv = -jf*t, Δω = -2 jf/R (with sig)
    ball.v.x -= jf*tx;
    ball.v.y -= jf*ty;
    ball.angVel += -2*jf/R;
  }
}

Wait sign of Δω. Earlier derived with friction opposing positive slip (s_contact>0) produced Δω=-2jf/R with jf>0. If s_contact negative, jf negative, then Δω positive. So unify: let jf = the scalar from formula (can be signed). Δv = -jf*t, Δω = -2*jf/R. Let me write:

const s_contact = srel + ball.angVel*R;
// desired jf:
let jf = s_contact/3;
const jfMax = mu*jn;
jf = clamp(jf, -jfMax, jfMax);
ball.v.x -= jf*tx;
ball.v.y -= jf*ty;
ball.angVel += -2*jf/R;

That handles both directions.

Hmm but Δω = -2*jf/R: since radius R used both as ball radius and lever arm, fine (same).

Apply the same to ball-ball collisions with appropriate R and I/m for both balls.

Ball-ball collision: standard impulse for two spheres with equal mass m (=1). Position resolution: separate by moving each half along normal by overlap. Velocity: relative along normal, restitution. Friction similar but with relative angular velocities of both.

For two balls equal radius r, I/m = 0.5 r².
At contact between ball a (center Ca) and b (Cb), normal n = (Cb-Ca)/d pointing from a to b. 
Relative velocity at contact: v_rel = (v_b + ω_b × (contact-Cb)) - (v_a + ω_a × (contact-Ca)).
contact = Ca + n*r.
Compute surface velocities similarly:
vel_a_at_contact = v_a + ω_a*(-(r*(-ny)), r*(-nx))? Let me reuse pattern. For ball a, r_vec from center to contact = n*r. spin surface vel = ω_a * (-(n.y*r), (n.x*r)) = ω_a*r*(-n.y, n.x). Let t2 = (-n.y, n.x) which is rotate n by +90°.
For ball b, r_vec = -n*r, spin vel = ω_b*r*(n.y,-n.x) = ω_b*r*t3 where t3=(n.y,-n.x) = -t2.

Normal relative velocity: vn = dot(v_b - v_a, n) for approach. If vn<0 approaching → respond with restitution e_bb.

This is getting elaborate. Given time constraints, let me implement ball-ball elastic with restitution + friction spin in a simplified but consistent way. Actually let me just do a clean two-body impulse with tangential friction including spin, similar to wall but with both masses.

Given the 15-minute budget, I'll implement it carefully but not over-engineer angular for ball-ball too perfectly; a reasonable epsilon is fine as long as numbers rotate and spin visible.

Let me write ball-ball:

function ballBall(a,b,dt){
  const dx=b.x-a.x, dy=b.y-a.y;
  const r= a.r;
  const d2=dx*dx+dy*dy;
  if(d2>= (2*r)*(2*r)) return;
  if(d2===0){ // coincident, nudge
     a.x += (Math.random()-0.5)*0.1; b.x += ... avoid
     return;
  }
  const d=Math.sqrt(d2);
  let nx=dx/d, ny=dy/d;
  // position
  const overlap = (2*r - d);
  a.x -= nx*overlap/2; a.y -= ny*overlap/2;
  b.x += nx*overlap/2; b.y += ny*overlap/2;
  // normal relative velocity (approach if negative)
  const vnx = b.v.x-a.v.x, vny=b.v.y-a.v.y;
  const vn = vnx*nx+vny*ny;
  const e=0.9; // ball-ball restitution
  if(vn<0){
    const jn = (1+e)*(-vn)/2; // divide by 2 because two equal masses, impulse split
    a.v.x -= jn*nx; a.v.y -= jn*ny;
    b.v.x += jn*nx; b.v.y += jn*ny;
    // tangential
    const tx=-ny, ty=nx; // perpendicular
    const srel = (vnx*tx+vny*ty); // relative surface motion along t
    // contact slip with spins: contact point of a at +n*r, of b at -n*r
    // surface vel a along t: v_a·t + ω_a*r (since spin at +n*r: ω_a*r*(-n.y,n.x)? let me compute
    // Let's compute contact slip using contact points.
    const ca_x = a.x + nx*r, ca_y=a.y+ny*r;
    const cb_x = b.x - nx*r, cb_y=b.y-ny*r;
    // surface velocity of a at ca: v_a + ω_a x (ca-aA center)
    const rax=ca_x-a.x, ray=ca_y-a.y; // = n*r
    const va1x = a.v.x + a.w*(-ray), va1y = a.v.y + a.w*rax;
    const rbx=cb_x-b.x, rby=cb_y-b.y; // = -n*r
    const vb1x = b.v.x + b.w*(-rby), vb1y = b.v.y + b.w*rbx;
    // relative surface velocity
    const srx=vb1x-va1x, sry=vb1y-va1y;
    const s = srx*tx+sry*ty; // slip along t
    const jn2 = (1+e)*(-vn); // total normal impulse per mass? Actually for friction use contact impulse.
    ...
  }
}

The spin coupling for ball-ball friction is more complex due to two bodies. Let me just apply a simple friction: reduce relative tangential slip via a restitution-like factor and transfer to spins.

Given time, I'll do: friction factor µ reduces relative tangential velocity; distribute angular update loosely. The visual requirement is "numbers rotate with balls" and "spin is visible". Ball-ball friction giving spin is a nice-to-have; even without ball-ball spin, rolling on walls gives spin. And with ball-ball, some angular coupling via walls is enough. But let me add basic ball-ball friction too.

Simplify ball-ball friction:
After normal impulse, compute relative tangential velocity rvt = (vnx*tx+vny*ty) with t=(-ny,nx). Let me compute contact-point relative tangential including spins (the s above). Then impulse jf_t reduces it. Apply:

The friction reduces slip. Let me set:
jf = µ_bb * jnMagnitude, sign opposing s.
Apply Δ to both balls' velocities along t and spins.

For equal masses and radii equal, distribute symmetrically: 
Let me use the formula solving no-slip for two bodies. Friction force f on b (and -f on a) applied at contact. 
For b: r_vec_b = -n*r. torque_b = r_b × F. For friction direction oppose slip along t.
For a: r_vec_a = n*r. torque_a = r_a × (-F).

Equal radii, equal mass, I/m=k*r² (k=0.5 disk). 
The no-slip constraint and impulse balance — this is doable but time. 

Let me implement it properly since math is straightforward:

Let t = (-ny, nx) (unit, ⊥ n).
Slip before: s = srx*tx+sry*ty (as computed). 
We apply friction that changes state. Let variables:
Δv_a = -jf*t (equal and opposite giving Δv_b=+jf*t) — actually linear impulses: on a: -jf*t, on b: +jf*t? The relative velocity we want to reduce; friction force on b is -f*t (opposing b's slip relative to a). Here slip s = relative velocity of b contact minus a contact along t. Friction acts to reduce |s|. Force on a = +f*t, on b = -f*t. (Force on b opposes its motion in +t.)
So Δv_a = +jf*t, Δv_b = -jf*t, where jf = f/m (magnitude, signed).
Δω_a: torque about a center from force +jf*t at contact (r_a=n*r): τ_a = r_a × F = n*r × (jf*t). Compute: (n.x*r)*(jf*t.y) - (n.y*r)*(jf*t.x) = jf*r*(n.x*t.y - n.y*t.x). n.x*t.y - n.y*t.x = n.x*(-n.x... wait t=( -ny, nx). n.x*t.y-n.y*t.x = n.x*nx - n.y*(-ny)=nx²+ny²=1. So τ_a = jf*r. Δω_a = τ_a/(k r²) = (jf*r)/(k r²) = jf/(k r).
Δω_b: force -jf*t at r_b=-n*r. τ_b = (-n*r) × (-jf*t) = (n*r)×(jf*t) = same as a = jf*r. Δω_b = jf/(k r).

Interesting both Δω same. k=0.5 → jf/(0.5 r)=2 jf / r. So Δω_a=Δω_b=2*jf/r.

Now slip after = slip_before + Δ. slip = (v_b + ω_b×r_b) - (v_a + ω_a×r_a), in t component.
Change in slip along t due to Δv_a,Δv_b,Δω:
Δslip_t = Δv_b_t - Δv_a_t + Δω_b*(r_b cross term t) - Δω_a*(r_a cross t).
Δv_b_t = -jf (dot (Δv_b,t)= -jf*(t·t)=-jf). Δv_a_t = +jf. So Δv_b_t-Δv_a_t = -2jf.
Spin contribution: for body with r_vec_fromcenter, spin surface vel = ω×r_vec where ω z-scalar. For b, r_b=-n*r. (ω_b×r_b) in t direction. Earlier we found spin surface velocity for ball = ω*r*t when r_vec=-n*r (ball bottom). Here r_b = -n*r, so spin surface vel of b = ω_b*r*t, component along t = ω_b*r. For a: r_a=n*r, spin vel = ω_a×r_a. Compute: ω_a×r_a: (-ω_a*r_a.y, ω_a*r_a.x)= (-ω_a*(n.y*r),ω_a*(n.x*r))=ω_a*r*(-ny,nx)=ω_a*r*t. Component t = ω_a*r.
So spin slip contribution = (ω_b*r) - (ω_a*r) = (ω_b-ω_a)*r.
Changes: Δ(ω_b-ω_a)= Δω_b-Δω_a = 0 (both equal). So spin change contributes 0 to slip along t! Interesting. So friction via spins doesn't directly change slip-relative... because equal radii both spin same way but that changes the constraint? Hmm.

Wait that means with symmetric equal radii, applying friction impulse changes linear slip by -2jf but spins adjust equally not affecting slip. That seems odd but mathematically the equal-radii symmetric case: torque changes both spins equally, but the slip reduction comes only from linear velocity change. So jf = s/2 to kill slip (from -2jf term). Then clamp by Coulomb.

OK so for ball-ball equal mass equal radius:
jf_desired = s/2. Clamp |jf|<= µ_bb*jn_per_impulse. 
Δv_a=+jf*t, Δv_b=-jf*t, Δω_a=2jf/r, Δω_b=2jf/r.

Good enough. Let me use that.

Actually let me define jn for friction normalization: jn_mag = (1+e)*(-vn) (total normal impulse, magnitude, >0 when approaching). µ_bb maybe 0.2.

Alright, let me also add a small angular damping and also transfer: when balls roll, angVel should relate. Fine.

Let me now also handle gravity direction and make sure balls settle at bottom inside rotating heptagon. Since the heptagon rotates and walls move, balls at bottom will ride the bottom wall, getting pushed/spun. When wall at bottom rotates down, the ball might get carried but gravity keeps it roughly bottom. Could the ball get sandwiched and pushed up along the wall requiring it to climb? The bottom vertex moves; as the polygon rotates, a vertex at bottom moves up the side, the ball (on floor) ... Actually the "floor" of the rotating polygon changes. Balls will continually be pushed by walls. They remain contained (we keep pushing inward). This should look lively. Good.

Potential instability: The moving walls push balls with wall velocity which can be large: ω*R = (2π/5)*300 ≈ 377 px/s. That's manageable. Balls hitting a fast-moving wall get kicked. Keeping substep small.

One more: When pushing ball outward after penetration using closest point, if the closest point is past the vertex (segment edge), the ball might be pushed outside the polygon if the normal points wrong. But n = (center-closest) direction, pointing into the interior region near that edge. Since closest is on the edge and is the nearest, the direction (center→closest) is perpendicular-ish to edge pointing outward (toward center/internal). Actually center is inside, closest on boundary, so center-closest points from boundary inward = interior. Good and correct.

But careful: the "outward normal" for pushing should point into interior. Since ball is inside (dist<R to that edge's interior? not necessarily, the ball near a vertex of a concave... polygon is convex). For a point inside a convex polygon, for each edge the signed distance (center - edge midpoint)·(edge inward normal) is positive (inside half-plane). The closest point's direction approximates inward normal. Good.

I'll rely on the closest-point approach. It's generally robust for convex shapes.

Let me also, for extra stability, after all iterations, re-clamp positions into the polygon region (if a ball center is outside polygon, push it back to nearest edge). But that only happens on severe failure; skip.

Let me write it all and test in browser.

Let me choose canvas: I'll make it responsive but fixed logical 800x800 with CSS scale to fit. Simpler: set size to a fixed square and center. I'll detect container width. Let me just do 800x800 fixed, and CSS max-width:100% with aspect ratio. Actually scaling canvas via CSS distorts pointer coords but we don't need pointer input except buttons. Fine.

Let me write the HTML.

Let me include:
- <style> dark background, centered layout, buttons.
- canvas #c
- control div with buttons Pause/Resume, Reset
- info div showing time
- <script>

Draw: clear, translate to center? I'll draw in world coords with canvas origin at 0,0 and center at O. I'll reserve top area maybe but keep heptagon centered within 800x800. Let me make canvas 860x820 with heptagon center ~ (430, 440) to allow buttons outside canvas in HTML (controls below). Simpler: controls in HTML above/below canvas; canvas 800x800, center (400,400).

Let me now also compute omega in radians per second: ω = 2π/5.

Rotation angle θ(t) = theta0 + ω*simTime.

Reset: store initial balls (positions, velocities, angles), and set simTime=0, θ=theta0, clear angVel.

Sim time increments only when not paused.

Let me code now.

For gravity sign: in canvas y increases downward, so gravity + in y (downward). G positive.

Let me write ball creation:

function makeInitialBalls(){
  const balls=[];
  const r=BALL_R;
  const spacing=BALL_R*2 + 0.3;
  const rowH = spacing*Math.sqrt(3)/2;
  let idx=0;
  // rows: [1,2,3,4,5,5]
  const rows=[1,2,3,4,5,5];
  for(let ri=0;ri<rows.length && idx<20;ri++){
    const c=rows[ri];
    const y=cx + (ri - (rows.length-1)/2)*rowH; // center cluster vertically
    for(let j=0;j<c && idx<20;j++){
      const x=cx + (j-(c-1)/2)*spacing + ((ri%2)? spacing/2:0);
      balls.push({x,y,vx:0,vy:0,ang:0,w:0,num:idx+1, r:BALL_R, color:colorFor(idx)});
      idx++;
    }
  }
  return balls;
}

Vertically center: cluster centered at O.y = cy. rows.length=6. (ri - 2.5) gives -2.5..2.5, centered. Good.

Colors: rainbow by index.

Let me write it all now and test.

For stability with the rotating wall and gravity, dt fixed 1/120 with accumulator, and 3 constraint iterations per step. 20 balls * 7 edges * 3 iter = 420 wall checks + 190 pairs*3=570 ball checks per physics step, at 120 steps/sec = ~118k checks/sec. Fine.

Actually doing 120 physics steps/sec plus render = 60fps. Accumulator approach: dt_step=1/120. Might do 2 steps per frame at 60fps. Fine.

Let me implement now.

Let me be careful about the pause: when paused, stop stepping simTime and physics but still render (so rotation frozen). Reset button restores.

Let me write the file.

I'll write angular damping: aw *= Math.exp(-angDamp*dt) with angDamp ~ 1.5. Also maybe slight, keep spins.

Also add rolling-to-linear? Not needed.

Let me also add small random... no keep deterministic for reset.

Now the rotation of number with ball: in draw, translate to ball pos, rotate by ball.ang, draw circle, draw number, restore. Since ang starts 0 and changes, number rotates. Good.

Also to make spin visible especially initial: initial angVel 0, but walls impart spin via friction quickly. Good.

Let me code.

Also velocity clamp: maxSpeed ~ 2000; if |v|>max clamp.

Let me write the file and then test in browser.

Let me also add retain: first physics after creation, balls fall under gravity to the bottom cluster; they will spread and hit walls. Nice.

Let me write.

I'll define constants:
const W=800,H=800;
const O={x:400,y:400};
const R=310; // heptagon circumradius (fits in 800, max extent 310 <400)
const BALL_R=26;
const DAMP=0.995; // per-step linear damping? maybe small
const EXTRA... 

Let me include slight linear air drag: v *= (1 - airDrag*dt), airDrag=0.3. Helps stability. Actually set small.

Let me finalize and write.

I need to also ensure balls don't overlap initially: spacing = 2r+0.3, yes non-overlapping.

Let me make sure BALL_R=26 with 310 circumradius gives inradius 310*cos(pi/7)=310*0.90097=279. Ball fits. Good.

Now write.

Let me also draw the heptagon: compute 7 vertices, stroke path. Maybe fill translucent, plus show dots at vertices, and a small rotation marker (line from center to one vertex) to visualize rotation. Good—helps confirm rotation.

Let me code the whole thing.

I'll now write index.html.

Physics step world function returns updated.

Let me write carefully the simulation loop using requestAnimationFrame with accumulator, clamping dt to avoid spiral (dt <= 0.05).

Let me trace the friction sign for wall once more with concrete: ball rests on flat bottom wall, wall not moving (theta fixed), n=(0,-1) pointing up (from wall into ball, interior up). Wait interior is upward (+... in canvas y down, interior above the bottom wall means -y? bottom wall is at bottom, interior above it, i.e., -y direction? Hmm canvas y down. Let default: n pointing from wall into interior. For bottom wall, interior is up = -y? No: bottom of screen is large y, above it (toward center) is smaller y. Ball sits on bottom wall, its center above (smaller y). normal from wall toward ball points up = -y direction. So n=(0,-1). 

t=(n.y,-n.x)=( -1,0) → t=(-1,0) i.e., -x direction. Hmm t could be either; our chosen t=(n.y,-n.x). For n=(0,-1): t=(-1,0). 

Ball rolling to the right (+x). Ball v_x>0. Wall v=0. rel_v=(vx,0). srel = relx*tx = vx*(-1) = -vx <0. s_contact = srel + w*r. 
Friction: if ball rolls right with w such that no slip... Let's see the physics: ball on floor moving right, to roll without slipping it spins clockwise (w<0) with w*r = -vx? Because surface contact velocity = v_ball + w*r*t. For rolling right: v_x>0, spin clockwise w<0. Contact surface vel = v_x + w*r*(t·x). t·x = -1 (t=(-1,0)). = v_x + w*r*(-1) = v_x - w*r. For no slip =0 → w = v_x/r >0?? That gives w>0. Contradiction. Let me recheck my spin surface velocity direction formula.

Earlier: r_vec = -n*r (contact offset from center toward wall). With n=(0,-1), r_vec = -n*r = -(0,-1)*r = (0,r) → contact is at center+(0,r) = below center = bottom point. Good. spin surface vel = w*(-r_vec.y, r_vec.x) = w*(-r, 0)=(-w*r,0). So for bottom point, spin surface velocity is in -x when w>0 (CCW). Component along t=(-1,0): dot((-w*r,0),(-1,0)) = w*r. So contact surface velocity along t = v_x-tangent + w*r. v_x along t: v_x*(t·x)=v_x*(-1) = -v_x. So s_contact = -v_x + w*r. Set 0 → w = v_x/r. For rolling right v_x>0 → w>0 = CCW. But physically rolling right on floor is clockwise. Sign mismatch because of canvas y-down/rotation orientation. Since our rotation is defined as +angle CCW in standard math but canvas y flips, angular velocity positive means... visually CCW in canvas is actually clockwise in math terms. Whatever—the numbers rotate, which is what matters visually. The physics consistency internally holds (friction drives s_contact→0). Fine.

So w = v_x/r positive means in canvas a ball rolling to the right spins with positive angular (which visually is counterclockwise in canvas? in canvas positive angle rotates from +x toward +y=down = clockwise visually). It's consistent no-slip. Good enough.

Friction: jf_desired = s_contact/3 = ( -v_x + w*r)/3. If the ball is dropped without spin (w=0) moving right v_x>0, s_contact = -v_x <0, jf = s_contact/3 <0. Δv = -jf*t = -jf*(-1,0) wait t=(-1,0), -jf*t with jf negative → -jf positive → -jf*t = positive... let me just: Δv = -jf*t. jf<0 → -jf>0, times t=(-1,0) → (-vx-ish) hmm. Let me just trust formula; it reduces slip.

n=(0,-1), t=(-1,0). Δv_a = -jf*t = -jf*(-1,0)=(jf,0). jf negative → Δv negative x → slows ball rightward. Good that reduces slip. Δω = -2*jf/R: jf<0 → Δω>0 (spin CCW canvas), matching rolling-right positive w. Good consistent.

Great, formula works.

Now let me write everything.

I'll set mu=0.4. Restitution e=0.5 for walls, e_bl=0.85 balls.

Let me also cap per-step so friction doesn't cause issues. Fine.

Writing now.

Let me make the canvas a bit larger R to look good: R=330, canvas 820x840? Let's keep 800x800, R=315.

Time to write.

I'll also add a subtle check: keep ball positions and if they exceed some radius from center (shouldn't), clamp.

Let me produce the file.

For rendering numbers, use a distinct color and white number with outline.

Let me write.

After writing, I'll test with browser tool: reload, screenshot, inspect console, click Reset, screenshot, interact. Let me proceed.

Let me write the full HTML now.

Let me take a screenshot to see. The browser tool says images not supported. Hmm. Let me try inspect to get DOM. Let me try screenshot anyway.

Actually images not supported. Let me check console for errors. No errors. Let me wait and inspect time progression and rotation to verify.

Let me take a screenshot attempt - model does not support images per output but let me try.

Images not supported by the model. I'll rely on DOM/JS inspection. Let me use browser inspect and evaluate via clicking and reading console. But I can't run custom JS easily? The browser tool has actions reload, inspect, click, etc. I need to verify physics. Let me at least verify the info counters progress and theta rotates, and that balls have velocities/separations.

Let me instrument by evaluating in the page. The browser tool doesn't expose an evaluate action directly except via inspect. Hmm. Let me check the motion action - maybe it returns frames. Let me try motion.

Actually, let me add temporary debugging via console.log from the page? That requires editing. Alternatively I can use the tool's "inspect" to see current DOM. 

The info line shows t and θ progressing - rotation works (θ changes).

To verify balls exist and move, let me inject a check by editing to log, or better, I can evaluate by using the "key"/"type"? Not for JS.

Let me use the browser motion action to see frames. Let me try.

Now I need to add debugReadout function inside the IIFE so it's accessible. Let me add it.

Good. avgP near center (415,439), start. Let me sample over time to see behavior during the fall and collisions: check that balls stay contained (minPairD >= ~54 means no overlap), speeds bounded, and avgP rises as it falls? Actually gravity falls down (y increases). Let me poll a few times.

Balls have fallen to the bottom (avgP y=543, x=285 shifted left). avgW 1.9 means spin present! minPairD 54.0 = no overlap exactly at 2r. Good. AvgV 191 bounded. Let me sample again to see continued dynamics and check no overlap blowup.

Stable. balls moving, spin avgW still present, minPairD 54.1 (>= 2r, good, no overlap). The rotating walls are continuously pushing balls. θ changed 278°.

Now let me test Pause/Resume: click pause, verify time stops and θ stops; then resume.

Let me click pause button. The button coordinates... Let me inspect DOM to find button location. Simpler: there's a click action with x,y. Let me find button positions. The DOM text order: pause button is after canvas. Let me estimate. Canvas 800x800 shown scaled; the browser viewport is 1200x800. The wrap is centered. Hard to know exact pixel coords for the button.

Better: I can just use keyboard? No. Let me click roughly. Actually the browser click uses viewport coordinates 1200x800. The page content: body flex centered. Canvas 800px wide scaled? Canvas is 800x800 but viewport 1200x800 so it may be scaled down by CSS? No, canvas has fixed width 800 attribute; CSS doesn't scale it (no max-width). So content is 800 wide + padding. Centered horizontally in 1200 → left edge ~ (1200-800)/2=200. Vertically, header + canvas + controls. Center heptagon.

The buttons: controls below canvas. Canvas top ~ center of screen minus ~ (800/2 + header + padding). Hard.

Let me instead just verify pause logic via editing? That's overkill. Let me try clicking. Buttons: pause is left, reset right, in a row centered, below canvas.

Let me estimate: total wrap height ~ 800 canvas + 16 padding*2 + header 20 + controls ~40 + info ~20 ≈ 900px tall. Vertically centered in 800 viewport → extends beyond viewport! 900>800. So canvas bottom is cut off. Hmm. The canvas might be partially below the fold. Buttons maybe below 800 → not clickable/visible.

This is a problem for usability. Let me reduce layout so everything fits in 800px viewport. The canvas 800px tall is too big combined with controls. Let me reduce canvas to ~560px and keep logical coordinate 800x800 by scaling? If I reduce canvas DOM size to 560 and keep internal 800, I'd need to scale ctx. Simpler: keep canvas at logical 800 but set its CSS width/height to 580px, and scale the drawing context by 580/800. Or make canvas element 580x580 and adjust internal constants. 

Actually simplest: reduce canvas attributes to 640x640 and reduce R_CIRC/BALL_R accordingly? That changes physics scale (lower resolution but fine). But keeping 800x800 logical with smaller rendered size is cleaner visually.

Let me set CSS to scale canvas: canvas element attribute width=800,height=800 but style width 560px, height auto. Then controls come below, total ~ height < 800. The heptagon (circumradius 320) in 800 logical → max extent 320 from center 400 → sphere 80..720, within 800. When scaled to 560px it fits. Good.

Let me add CSS to canvas: `width:min(560px, 90vw); height:auto;`. And ensure body can scroll vertically too in case. Flex center but allow overflow.

Let me check viewport 1200x800 given by browser. Actually the browser viewport is 1200x800 per tool description. So I want total page height ≤ 800. 

Set canvas rendered width ~520px. Height auto = 520. Header ~24, padding ~32, controls ~40, info ~16 → ~632 total, plus h1 margin. Fits.

Let me update CSS.

Then test pause/reset. Let me also make layout not overflow.

Now let me test pause by clicking. I need button coordinates. But I removed info debug? No I kept it but edited the info text; the debugReadout is still appended. Info line still shows. Good.

Let me find button pixel coords. The wrap now: h1 20 + padding, canvas width 520 scaled, controls, info. Let me click the Pause button. I need to determine its position. 

Let me use browser motion? No. Let me just try clicking near center-bottom. Actually I can compute: page centered. Wrap width = max(canvas 520, controls width ~ 200) + padding 32 → ~ 552. Height: padding16*2=32 + h1 ~24 + margin10 + canvas 520 + margin12 + controls ~37 + info ~16 ≈ 651. Top ~ (800-651)/2 ≈ 74. Canvas top ≈ 74+16+24+10 = 124, canvas height 520 → canvas bottom ≈ 644. controls at ≈656, center vertically ~673, horizontal center 600. Buttons row centered in 552-wide wrap: pause at left ~ 600 - (wrapcontent/2). Let me just click x=600 (center of viewport) y=675 for pause, and reset slightly right.

Actually pause and reset in a row gap 10. Pause left of reset. Center row at wrap center. Pause center x ≈ 600-55=545? Let me click around. Let me click x=560,y=675.

Let me first inspect to confirm the pause button is visible/on-screen. inspect DOM earlier listed buttons. Let me get bounding rect via... can't. Just try clicking and check info changes to paused.

Time frozen at t=8.87, θ=278°. Good. Now resume by clicking pause (Resume) button same location.

The click toggled back to Pause (so it resumed) but the info still says paused? Wait the click result shows button back to "Pause" and info still says paused... but that's the DOM snapshot at click time before the next frame updates. Let me inspect again after a moment.

Resumed and running (t advanced to 14.63, θ 333°). 

Now test Reset. Need to click Reset button. It's to the right of Pause. Let me click x~640, y~675. Let me first test Reset: it should restore balls to near center (avgP ~415,438) and θ to initial.

But note: reset restores saved snapshot taken at init when simTime=0, theta=THETA0=0. However, when user paused mid-run and reset, saved should be the original initial state. But I only set `saved` once at init. If user presses Reset, it restores original init state. Good, matches "reset restores initial ball arrangement and wall orientation". 

However there's a subtle issue: after Reset, the saved snapshot has paused=false always (original). That's fine—the task only requires restore arrangement and orientation. Should Reset also keep current pause state? Not required. But arguably restart should run. Currently after reset, it resumes running regardless of prior pause. Acceptable.

Let me click Reset and verify.

The click at 645,675 hit... nothing changed (still t=21.41). Maybe my coordinates off. The info didn't reset to t small. So Reset button click missed. Let me find the actual button coordinates. 

Let me use browser to get element positions. There's no evaluate. But I can temporarily log coordinates? Alternatively, I can add a debug tool: output bounding rects into the info DOM. Let me add to debugReadout the rect of reset button. Actually better, I'll locate buttons via clicking methodically.

Let me estimate again: viewport 1200x800. Content centered. Wrap width: canvas min(520, 0.84*1200=1008)=520. Width 520 + padding 32 = 552 wrap. Centered → left edge x=(1200-552)/2=324, right edge 876. Controls row centered at x=600. Buttons: two buttons inline with gap 10. "Pause" width ~ 70px, "Reset" ~70px. Total ~150. So pause occupies x from 600-80 to 600-10 → around 520-590. reset from 610 to 680. My click 560 hit pause (correct). My reset at 645 should hit. But it didn't reset. 

Wait, maybe the click at 645 landed but the page had just been clicked twice (resume click earlier also at 560). Let me re-try reset. Actually the inspect after shows t=21.41 running with same avg, meaning reset didn't trigger. Possibly 645,675 was within reset button but timing: the click handler restores saved which has pauseLabel update. If it had worked, t would go to ~0 and θ ~0. It didn't. So missed.

Let me try clicking reset at different y. Actually maybe the buttons vertical location is different than 675. Let me recheck: after resizing canvas to 520, page height smaller, content centered higher. wrap height: padding 32 + h1 line ~ 24+10 + canvas 520 + controls margin 12 + button ~37 + info. Total ~ 635. top = (800-635)/2 ≈ 82. Controls y center ≈ 82+16+34+10+520+12+18 ≈ 692. So buttons around y=690, not 675. My pause click at 675 still toggled because button height ~37 spans y≈674-712. 675 top edge hit? It worked for pause. Reset at 645,675 should also be within vertical range but maybe x: reset center more like 640? I clicked 645,675 — top edge of button. Buffers: button top ~674. 675 is 1px in. Might be borderline. Let me click reset at 645, 695.

Let me verify positions more robustly by adding rect info to debug output. That's the reliable approach. Let me add to debugReadout the reset button rect and pause rect.

The rects aren't showing because "BTN" appended via rects then... wait I return rects at end but the caller appends ' | ' + debugReadout(). The rects should be in output. But output shows avgP... The info text got truncated? The console shows dom text but maybe my ' | ' join where debugReadout returns rects + spaces + the ball data. Output shows only "minPairD" and no BTN. That means debugReadout returned earlier? I moved the return rects to inside before ball computation, and it returns rects only when !balls.length. Otherwise it continues and returns the long string which includes rects + ball data. So rects should be included. But they're not shown. 

Maybe getBoundingClientRect during early frame returns 0s and Math.round → but still prints "BTNreset(0,0,...)". It's not showing at all though. Possibly the debugReadout string is present but the info DOM text is truncated in the tool output? The console output shows a truncated line "dom": which is the full textContent but seems cut at 2r=54. Actually maybe it IS truncated by the tool for length. The textContent is long now.

Hmm, the output shows up to "(minPairD 54.0 (2r=54)" with no closing — suggests truncation of the reported dom string. So rects ARE there but truncated in the tool's view. 

Let me instead shorten: put rects at the FRONT? No, they're at front of debugReadout but debugReadout appears at very end of info string after ' | '. So rects are near end → truncated.

Let me instead output button rects via a separate DOM element placed after info, or reorder so debugReadout rects come... The info string = base + ' | ' + (rects + balldata). Truncation cuts the tail. Let me move rect info to the front by putting it in base? Simpler: create a second div for debug below, but that also may be truncated.

Alternative: just probe buttons by clicking at several offsets until reset triggers. Let me do methodical clicking. I'll click reset at (645, 700), (645,690), (600,700), etc. Actually simplest: after reload, quickly (within a second) the page layout is stable. Button rects from estimate: reset near right of center. Let me just click around right side at y 690.

But note after reload the sim runs; I can still reset anytime.

Let me click reset at x=645, y=692.

Not reset (t=14.73, not near 0). So 645,690 still missed. My estimate of button y is off. Let me think: page after canvas resize — total height maybe ~640, top ~80. h1 then canvas. Controls below canvas. Let me reconsider vertical: 
- padding top 16
- h1 20px + bottom margin 10 → 30ish, height ~24
- canvas 520
- margin-top controls 12
- button height ~ 30-37
- info 16 + padding bottom 16

Sum ~ 16+24+520+12+33+16+16 ≈ 637. top=(800-637)/2≈81. 
Controls center y ≈ 81 +16+24+520+12 + 16 ≈ 669. So y~669 or 670. My clicks at 675 worked for pause (toggling), consistent. Reset at 690,692 should be within button (button spans ~652-686). Maybe reset x at 645 is between buttons? Let me check x placement:
wrap width 552, centered → left 324, center 600. Controls row centered at 600. Pause button ~ width 66 (text "Pause" + padding 36). Reset ~ 66. Gap 10. Total ~142. So pause spans x[529,595], gap[595,605], reset[605,671]. Center of reset ≈ 638. My x=645 within. So y must be wrong. At y=690 if button spans 652-686, 690 is below. Let me try y=670, x=640.

Let me click 640,670.

Reset worked now! t=0.02, θ=1°, avgP(415,436) back near center, avgW 0. This restores the initial arrangement and wall orientation (θ ~0). 

minPairD 53.6 slightly less than 54 initially due to the 0.4 overlap? No, spacing=2r+0.4 so pairs should be ≥54.4 initially. minPairD 53.6 at t=0.02 after 0.02s of falling — gravity has slightly settled, and iterations might have resolved residuals. It's ≥ ~53.6 >... well <2r=54 slightly (53.6<54). Minor initial overlap of <1px. Actually spacing between horizontally adjacent = 2r+0.4 = 54.4, no overlap. But 53.6 < 54? That's odd. Wait maybe some diagonal neighbors are closer. In hex lattice, nearest neighbor spacing 54.4; rows offset by 0.5 → nearest = spacing 54.4. So min should be 54.4. But 53.6 measured. Perhaps the reset snapshot balls were computed with spacing=2r+0.4=2*27+0.4=54.4. So minPairD ~54.4. But shown 53.6. Slight discrepancy; maybe X coordinate rounding? In makeInitialBalls x uses (j-(c-1)/2)*spacing, spacing=54.4. For a given row that's exact 54.4. Diagonals: distance between row i and i+1, offset by 0.5*spacing=27.2, vertical rowH=47.1. dist=sqrt(27.2²+47.1²)=sqrt(740+2218)=sqrt(2958)=54.4. So 54.4. Hmm but measured 53.6.

Anyway it's tiny (<1px) and immediately resolved by collisions; not a real overlap issue given gravity settles them. Fine.

Actually wait, at t=0.02 the balls have moved by gravity; avgV 17. minPairD 53.6 means one pair 0.4 px closer than touching. This happened because gravity moved upper ball down reducing gap slightly. Negligible; solver handles.

Everything works:
- 20 balls numbered 1-20 ✓ (present)
- initial compact non-overlapping arrangement near center, falls under gravity ✓
- heptagon rotates at 1 rev/5s: θ went from 0 to 340 over ~14s ≈ (340/360)*5? Actually OMEGA=2π/5, at t=14.73 θ should be 14.73/5*360=1060° → mod360=340. ✓ matches! Good rotation rate correct.
- collision detection, moving wall velocity accounted (friction with spin), damping ✓
- angular motion & numbers rotate (avgW nonzero) ✓
- stable: balls contained, minPairD ≥ ~54, speeds ~200 bounded ✓
- Pause/Resume ✓, Reset ✓ restores arrangement (avgP near center, θ~0) and wall orientation ✓

Now let me clean up the debug code (remove debugReadout and its usage, and BALL_LAST_ANG is fine to keep). Let me remove the debug additions to keep file clean. Actually keeping some debug isn't harmful, but better to clean it. Let me remove the debugReadout function and the appended ' | ' + debugReadout() and the button rect part.

Also I should verify the spin / numbers visible — can't see images. But logic ensures ang changes; draw rotates by b.ang. Good enough; I described it.

Let me remove debug code.

Let me read the area around the debugReadout function to get exact text.

My edit newText expected `if (!balls.length) return '';` but it's gone since previous edit replaced it. Let me remove the whole function block by matching lines 332-345.

Clean and running, no errors. Final check done. Let me verify index.html exists.